Engineering Tribology 2E Episode 6 pdf

... velocity [m/s]; 240 ENGINEERING TRIBOLOGY Computer Program for the Analysis of Grooved 360 ° Journal Bearings The example of a computer program ‘GROOVE’ for analysing of a grooved 360 ° journal bearing ... large. It can be seen from Figure 5.25 that for a 36 subtended groove angle and 0 .6 relative groove length, the dimensionless flow is about 6. 8. Assuming that the bearing entrainin...

Ngày tải lên: 13/08/2014, 16:21

... of Chemical Engineering Data, Vol. 6, 1 961 , pp. 112-118. 26 G. Salomon, A. Begelinger and A.W.J. De Gee, Friction Properties of Phthalocyanine Pigments, Wear, Vol. 10, 1 967 , pp. 383-3 96. 27 P.A. ... Lubrication Engineering, Vol. 22, 1 966 , pp. 262 - 269 . 31 W.E. Jamison and S.L. Cosgrove, Friction Characteristics of Transition-Metal Disulfides and Di-Selenides, ASLE Transactions...

Ngày tải lên: 13/08/2014, 16:21

... Strand 13 X 6/ 6 X 19 180 Grade 110 Grade MBL kN 8X 0. 565 d X 0 .61 3 8 X 0.851 8 X 0.500 8 X 0. 368 d X 0.809 Wt kg/100 rn d X 0.407 d X 0.413 8 X 0. 563 8 X 0.550 ... 50-100 0.035 0.035 100-150 0. 06 0. 06 ~ ~~ ~ ~ ~~~ Diesel 75-1 75 0. 06 0.04 175-250 0.08 0. 06 25woo 0.10 0.08 40C -60 0 0.15 0.13 Over 60 0 0.15 0.13...

Ngày tải lên: 13/08/2014, 17:20

... polymers 66 2 Wear and friction of ceramic matrix composites 66 2 16. 5 Summary 66 3 References 66 3 APPENDIX 66 9 Introduction 66 9 A.1 User friendly interface 66 9 A.2 Program ‘VISCOSITY’ 67 1 CONTENTS ... Perfluoropolyalkylethers 63 3.4 Emulsions and aqueous lubricants 65 Manufacturing of emulsions 65 Characteristics 65 Applications 66 3.5 Greases 66 Manufacturing of g...

Ngày tải lên: 13/08/2014, 16:21

... [0.0395 × (15.14 − 6. 69) 2 − 2 .66 ] × (1 − 273/373) − 0.303 × 15.14 − 0.0241 × (17 .60 − 6. 69) 2 + 5.731 = (2.8204 − 2 .66 ) × 0. 268 1 − 4.5874 − 2. 868 6 + 5.731 = − 1 .68 20 C 0 = 0.1 860 Which means ... 3.52 ISO VG 5 4.14 4 .6 5. 06 ISO VG 7 6. 12 6. 8 7.48 ISO VG 10 9.00 10 11.0 ISO VG 15 13.5 15 16. 5 ISO VG 22 19.8 22 24.2 ISO VG 32 28.8 32 35.2 ISO VG 46 41.4 46 50 .6...

Ngày tải lên: 13/08/2014, 16:21

... 32, 1 968 , pp. 155- 161 . 26 D. Godfrey, Friction of Greases and Grease Components During Boundary Lubrication, ASLE Transactions, Vol. 7, 1 964 , pp. 24-31. 27 M.H. Jones and D. Scott, Industrial Tribology, ... Engrs., Vol. 182, Pt. 3A, 1 967 -1 968 , pp. 585- 593. 32 A.G. Papay, Oil-Soluble Friction Reducers, Theory and Application, Lubrication Engineering, Vol. 39, 1983, pp. 419-4 26....

Ngày tải lên: 13/08/2014, 16:21

... 1.4 0.4 06 1 .6 0.399 1.8 0.393 2.0 0.387 2.2 0.381 2.4 0.3 76 2 .6 0.371 2.8 0. 366  3.0 1.2 1.4 1 .6 1.8 2.0 2.2 2.4 2 .6 2.8 3.0 3.2 3.4 3 .6 3.8 4.0 X B K h 1 h 0 Pad ... equation: 1 36 ENGINEERING TRIBOLOGY TABLE 4.2 Pivot position for various ‘K’ ratios. 0 0 1 0.482 0.2 0. 466  0.4 0.453 0 .6 0...

Ngày tải lên: 13/08/2014, 16:21

... W [N] 2.7 861 × 10 1 -1.100 2. 460 2.512 0. 563 0.528 -1.090 -0.383 - - - 1.385 - 2 H [W] 3.9307× 10 3 -0.7 06 1.577 0.477 2.240 1.287 0.249 -0.204 - 1 .324 - - 3 ε 1. 266 6× 10 -2 0.5 36 1.120 -1.050 ... Vol. 181, Pt. 3B, 1 966 -1 967 , pp. 45-54. 31 J.R. Stokley and R.R. Donaldson, Misalignment Effects in 180° Partial Journal Bearings, ASLE Transactions, Vol. 12, 1 969 , pp. 2 16- 2 26....

Ngày tải lên: 13/08/2014, 16:21

... ASME, Journal of Basic Engineering, Vol. 87, 1 965 , pp. 163 - 169 . 8 H. Opitz, Pressure Pad Bearings, Conf. Lubrication and Wear, Fundamentals and Application to Design, London, 1 967 , Proc. Inst. Mech. ... LUBRICATION 259 Boundary conditions from Figure 6. 1 are: p = 0 at r = R Substituting into equation (6. 2), yields the constant ‘C’: C = 6 Q πh 3 lnR (6. 3) Hence the pressure dis...

Ngày tải lên: 13/08/2014, 16:21

... is given by [15]: 3 16 ENGINEERING TRIBOLOGY Ellipticity parameter k = 1.25 -2 -1 0 1 x = x b 2.5 6 1.25 Ellipticity parameter k = 6 5 × 10 6 10 × 10 6 15 × 10 6 20 × 10 6 0 0.0005 0.0010 0.0015 Dimensionless film ... = 1 R x + 1 R ax 1 R bx = 1 10 × 10 −3 + 1 15 × 10 −3 = 166 .67 ⇒ R x = 6 × 10 −3 [m] = 1 R y + 1 R ay 1 R by = 1 10 × 10 −3 + 1 15 × 10 −3 = 166 .67 ⇒...

Ngày tải lên: 13/08/2014, 16:21